# How do you solve 64x^3+27=0?#?

Sep 29, 2015

Factor $64 {x}^{3} + 27 = \left(4 x + 3\right) \left(16 {x}^{2} - 12 x + 9\right)$

Hence $64 {x}^{3} + 27 = 0$ when $4 x + 3 = 0$, that is when $x = - \frac{3}{4}$

#### Explanation:

In general ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Putting $a = 4 x$ and $b = 3$ we find:

$64 {x}^{3} + 27 = {\left(4 x\right)}^{3} + {3}^{3} = \left(4 x + 3\right) \left({\left(4 x\right)}^{2} - \left(4 x\right) \cdot 3 + {3}^{2}\right)$

$= \left(4 x + 3\right) \left(16 {x}^{2} - 12 x + 9\right)$

The remaining quadratic factor has no linear factors with Real coefficients.

You can tell that because its discriminant is negative:

$16 {x}^{2} - 12 x + 9$ is of the form $a {x}^{2} + b x + c$ with $a = 16$, $b = - 12$ and $c = 9$.

Its discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 12\right)}^{2} - \left(4 \times 16 \times 9\right) = 144 - 576$

$= - 432$

In fact, $16 {x}^{2} - 12 x + 9 = \left(4 x + 3 \omega\right) \left(4 x + 3 {\omega}^{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive cube root of unity.

$\omega$ is used when finding all the roots of a general cubic equation.

Sep 30, 2015

$x = - \frac{3}{4}$

#### Explanation:

(As an alternate method)
$64 {x}^{3} + 27 = 0$

$\Rightarrow {\left(4 x\right)}^{3} = {\left(- 3\right)}^{3}$

$\Rightarrow 4 x = - 3$

$\Rightarrow x = - \frac{3}{4}$